Cyclotron. A cyclotron is a type of particle accelerator. These high-energy part
ID: 1623452 • Letter: C
Question
Cyclotron. A cyclotron is a type of particle accelerator. These high-energy particles are also used in nuclear research and in hospitals to obtain radioactive preparations for medical and diagnostic purposes. The cyclotron is made up of two semi-cylinder's conductors that are opened on the straight side. These D-shaped conductors are made of a non-ferromagnetic material (e.g., copper) and are called "dees". The "dees" are placed in a homogenous magnetic field with a magnetic field that is perpendicular to the conductors, as shown in the figure. If an alternating voltage is applied to the dee conductors, and a charged particle enters near the center of the cyclotron (i.e., at point Z in ithe figure), it is accelerated and gains a velocity that runs perpendicular to the magnetic field inside of one of the dees. As we have seen, a magnetic field forces the particles to travel in a circular path. However, the voltage between the "dees" linearly accelerates the charged particles as they cross the gap, increasing their kinetic energy. This effect increases the radius of the circle and so the path is spiral. The high-speed particles exit the cyclotron in the moment when the radius of the path reaches the radius of the cyclotron. Consider when deuterons, the nucleii of heavy hydrogen, are accelerated in a cyclotron. You can assume that they start with essentially no kinetic energy and that everything stays non-relativistic. (a) Determine the frequency of the voltage source if the value of magnetic field strength in the cyclotron is 1.5 T and the mass of a deuteron is 3.3 times 10^-27 kg. (b) Determine the necessary cyclotron radius if we want the deuterons to exit the cyclotron with a kinetic energy of 16 MeV (1 eV = 1.602 times 10^-19 Joules) (c) How many times does the deuteron cross between the "dees", if the voltage between them is 50 kV?Explanation / Answer
frequency = f
then time period of each osscilation = T
then period of half osscialtion = T/2 = 1/2f
magnetic force on particle of mass ma anc harge q = qvB
ce ntripital force = mv^2/r
then mv^2/r = qvB
mv/r = qB
v = rqB/m
but v = 2*pi*r/T
so, v = 2*pi*rf = rqB/m
f = qB/m*2pi
a) q = 1.6*10^-19
B = 1.5 T
m = 3.3*10^-27 kg
f = 11580775.91 Hz
b) Max V = RqB/m ( where R = rmax)
Vmax = R*q*B/m
KE max = 0.5mv^2
r MAX = sqroot(KE max*2* m)/qB = 0.5419 m (KE max = 16 MeV = 16*10^6*1.602*10^-19 J)
c) V = 50 kV
Final KE = 1.6002*1.6*10^-19*10^6 J
evergy gained in every crossing = qV
so, n = 1.6002*16*10^-19*10^6 /qV = 320.04 turns
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